Phenomenon 1, 2 valent metals react with Cl2 and exothermic more, and 3 and 4 valent metals react wi

It is mainly a matter of lattice energy.

1. The lattice energy of +2 valence metal combined with Cl- is particularly large, while the lattice energy of O2- combined with O2- is not large (especially for +1 valence metal).

3,+4 valent metals are the opposite.

The amount of heat released is related to the stability of the product. The more stable the product, the more heat is released.

Obviously, due to the strong polarization ability of metal ions, it is covalent with chlorine compounds, so many metal chlorides are covalent compounds, and the electrons are not transferred completely, and the energy released is of course less pulled. In addition, because chlorine is 1 valence, there is a lot of chlorine distributed around the metal, and the repulsion force between them also increases the energy, and the reaction heat release is of course small. Close to 1:

1 of the ** ionic compounds, the reaction exothermic is very high such as magnesium oxide

Chlorine is a chlorine-chlorine bond cl-cl bond has only one covalent bond, oxygen is an O=O bond, and the bond energy is larger than the cl-cl bond, which is the reason why the oxidation of chlorine is stronger than that of oxygen.

As for you saying that +1, +2 valence metals react with Cl2 to exothermic more, and +3, +4 valence metals react with O2 to exothermic more.

It's not just about the bond energy of chlorine and oxygen, it's about the properties of the metal, and it's hard for me to say, the teacher didn't talk about these things, you're in competition.

The problem doesn't seem to be clear!

Are you just one metal.

Through the chemical equation, it can be known that 1 mol HCl can be oxidized to produce 1 mol Cl2.

Therefore, the mass of Cl2 produced by the oxidation of G HCl is:

g HCl x (1 mol Cl2 1 molar HCl) x ( g Cl2 1 molar Cl2) = g Cl2

So the mass of the generated cl2 is and guess friend g.

MnO2+4HCl(concentrated)=MnCl2+Cl2 From the equation, it can be seen that 2HCl in 4HCl is formed into Cl2 by the oxygen type, and the HCl mass is known.

then the material is.

cl2 mass =

x +cl2== xcl2

1 x+x+x==64, that is, relative to the front of the pants, Henghu does atomic weight, and the regret refers to 64 is copper.

Option A should be: the energy of 12GC and 16g of oxygen must be higher than the energy of 28GCo.

A has elemental substances but no elemental reactants, so it is not a displacement reaction, so A is wrong;

b : Cl2 +2KBR=2KCl+BR2 The oxidant is chlorine, and the oxidation product is bromine, so the oxidation of chlorine is greater than that of bromine, KCLO3 +6HCl=3Cl2 +KCL+3H2 o The oxidant is potassium chlorate, the oxidation product is chlorine, the oxidation of potassium chlorate is greater than that of chlorine, 2KBRO3 +Cl2 =BR2 +2KCLO3 The oxidant is potassium bromate, and the oxidation product is potassium chlorate, so the oxidation of potassium bromate is greater than that of potassium chlorate, in short, The order of oxidation strength is KBRO3 KCLO3 Cl2 BR2, so B is correct;

c If (stp) is obtained in the reaction, then the number of electrons transferred = 1 (5-0) n

a mol = , so c is wrong;

d The amount of electrons obtained by 1mol of oxidant in the reaction 2kbro3 +cl2 br2 +2kclo3 is 1mol (5-0) 2=10mol, so d is wrong;

Therefore, choose B

Solution: Let the volume of concentrated hydrochloric acid to participate in the anti-draft be x, and the volume of chlorine gas generated by the spike diameter of y manganese dioxide = 30g*

mNO2+4HCl(Nongchai imitation town) = mnCl2+Cl2 +2H2O12mol L*x Y

x = y = A: (1) Calculate the volume of concentrated hydrochloric acid participating in the reaction of 88 ml (2) Generate the volume of CI2 liters.

When 2Molo2 is reduced, the heat of a kj can be obtained: 4HCl(g)+O2(g)=2Cl2(g)+2H2O(unknown) δH= A 2 kJ mol; Let the 1 mol H-O bond be broken to absorb XKJ, and the 1 mol H-Cl bond should be broken to absorb YKJ, H=4 Y+B (2C+4X)= A2

x－y=(a+2b-4c)/8

Answer: C Reason: O2 loses 4e 4NO2 gains 4e

Therefore, there is no electron gain or loss for metals.